1.

\(\cfrac{1+\sqrt2}{3-2\sqrt2}\) = a + b√2 then (a, b) = …………………A) (7, 5) B) (-7, 5) C (3, 5) D) (-7, -5)

Answer»

Correct option is (A) (7, 5)

Given that \(\frac{1+\sqrt2}{3-2\sqrt2}=a+b\sqrt2\)

\(\Rightarrow\) \(\frac{1+\sqrt2}{3-2\sqrt2}\times\frac{3+2\sqrt2}{3+2\sqrt2}=a+b\sqrt2\)

\(\Rightarrow\) \(\frac{(1+\sqrt2)\,(3+2\sqrt2)}{3^2-(2\sqrt2)^2}\) \(=a+b\sqrt2\)

\(\Rightarrow\) \(\frac{3+3\sqrt2+2\sqrt2+4}{9-8}\) \(=a+b\sqrt2\)

\(\Rightarrow\) \(a+b\sqrt2=7+5\sqrt2\)

\(\therefore\) a = 7 & b = 5   (By comparing rational and irrational parts of both equal real number)

\(\therefore\) (a, b) = (7, 5)

Correct option is (A) (7, 5)



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